An in depth analysis of the maths behind Peukert's Equation (Peukert's Law)

You will remember from here that Peukert's Equation is usually quoted as being T=C/Iⁿ. You will also remember we stated that whilst this equation is correct, it cannot be used with the usual battery data that is available. This is because the only battery data available is usually something like the 20 hour discharge rate and Peukert's exponent. For this reason we modified the equation to T=C(C/R)^n-1/Iⁿ by introducing another term [(C/R)^n-1] that takes into account the battery hour rating and capacity.

This allows the equation to be used correctly when given the 20 hour rate, or the 10 hour rate, or indeed any other rate. As long as you have available the amp hour capacity at a certain rate (eg 100Ahrs @ 20hr rate) and Peukert's exponent for the battery type, this formula can be used to predict run times at other discharge rates.

However many batteries are sold that do not make Peukert's exponent available. What they often show is a table giving different run times at different discharge rates, or a graph of discharge rates against run times which will look similar to the graph produced on our Peukert calculator spreadhseet.

What I intend to do here is run through the maths to show how Peukert's exponent can be calculated from these graphs or tables, or by running two discharge tests at two different discharge rates.

Be warned that the maths is rather involved.

Firstly a word of explantion. There are two ways to consider how Peukert's Equation affects available battery capacity:-

The first is to consider that the battery has a fixed capacity (say 100Ahrs), and increasing the current drawn from it takes progressively more power from it according to the exponent applied to the discharge current Iⁿ where "n" is Peukert's exponent. In this method, drawing 10 amps from the battery would be calculated as assuming that the actual discharge current is 10ⁿ.

The second method is to assume that the actual current remains the same but that the battery capacity shrinks in accodance with the exponent Iⁿ.

Both methods use exactly the same maths and give exactly the same results for run times at different discharge rates. The only difference is in the wording of the explanation.

We use the former method. The method of a fixed capacity battery and varying discharge currents according to Peukert's exponent.

We will start with some definitions.

(i) C = "nominal" battery capacity. ie the number written on the battery or data sheet eg 100Ahrs
(ii) R = "hour rating" written on the battery or data sheet e.g. 20 hour
(iii) I = the "nominal" current at the given rate (ie C/R). i.e. how much current the battery can provide for the hour rating R. In this example 100/20 = 5 amps
(iv) n = Peukert's exponent eg 1.3
(v) I_p = the "Peukert" current. The equivalent current that the discharge will remove from the quoted battery capacity ie Iⁿ
(vi) C_p = the "Peukert" capacity (RIⁿ). This will be explained below

The equation T=C(C/R)^n-1/Iⁿ works given a discharge rate, hour rating and Peukert's exponent but it doesn't explain why it is a different equation to the one normally quoted of T=C/Iⁿ.

As explained in the original simple explanation of Peukert's Equation the original equation of T=C/Iⁿ does not quote any specific battery capacity rating. And if we try to use this equation quoting say the 20 hour rate, or the 100 hour rate, we will get different and conflicting results.

Let's run the above figures through the modified equation.

T=C(C/R)^n-1/Iⁿ

T=100(100/20)^0.3/5^1.3 = 20 hours.

This we know to be correct because the battery spec is 100Ahrs at 20 hours which gives us 5 amps for 20 hours.

If we now run three times this current (ie 15A) through the equation we get:-

T=100(100/20)^0.3/15^1.3 = 4.794 hours.

So we can now say that this battery can provide 15 amps for 4.794 hours. 4.794hrs * 15A = 71.92Ahrs @ 4.794 hour rating. This is just as valid as the rating of 100Ahrs @ 20 hours for a battery with a Peukertt's exponent of 1.3.

If we now put these numbers into the equation (ie defining the battery as being 71.92Ahrs at the 4.794 hour rate) with a discharge rate of 5 amps we get:-

T=71.92(71.92/4.794)^0.3/5^1.3 = 20 hours

So clearly this still works.

If we compute the capacity and the added term in the equation [(C/R)^n-1] on its own we get 100(100/20)^0.3 = 162Ahrs

If we compute the other new battery spec that we calculated above we get 71.92(71.92/4.794)^0.3 = 162Ahrs

In both cases we get 162Ahrs. This is the Peukert capacity shown in (vi) in the definitions above. And this will work whatever capacity/hour rating we choose. The answer will always be the same.

The reason is very simple. It is that the original Peukert's equation operates from the starting point that the battery capacity is the total amp hours that can be drawn from the battery at a discharge rate of 1 amp. Batteries are never specified this way so the extra term (C/R)^n-1 corrects the given capacity specification to match that at 1 amp current draw.

The mathematical proof if this is (from the definitions above):-

Since I=C/R and I_p=Iⁿ then I_p=(C/R)ⁿ

and C_p=RI therefore, C_p = (C/R)ⁿR = Cⁿ/R^n-1 = C(C/R)^n-1

Note the final term here: C(C/R)^n-1 which is how the capacity was defined in the modifed equation.

So now we find that this particular battery has an actual Peukert capacity (I_p) of 162Ahrs This is now a capacity that can be used with the normal Peukert's equation of T=C/Iⁿ

Let's try the two discharge currents that we worked through above:-

Discharge at 5 amps we get T=162/5^1.3 = 20 hours which we know to be correct.

Discharge at 15 amps we get T=162/15^1.3 = 4.79 hours which, again, we know to be correct.

Using the original version of Peukert's equation (after having corrected the battery capacity as shown above) you would find that currents higher than 1A would produce shorter run times than a simple amps*time calculation would indicate. Currents lower than 1A would produce run times longer than a simple amps*time calculation would indicate. Let's call this the "swap around" point.

When using our modified equation [ie T=C(C/R)^n-1/Iⁿ] you would find that this "swap around" point would be at the discharge current that corresponds to the battery capacity divided by the hour rating ie C/R. For this reason the Peukert corrected amps (I_p) will be different in each case but the final run times calculated with either equation will be identical because the battery capacity has been quoted differently in each case.

Hopefully that has now cleared up why Peukert's equation (the normal one) cannot be used with normal battery data. One can either use the modified version of the equation [T=C(C/R)^n-1/Iⁿ which inherently corrects for the battery being specified at a discharge rate other than 1A] or one can convert the specified battery capacity to the 1A discharge rate and then use the normal version T=C/Iⁿ.

Now where is all this leading?

Well, most often the battery data, the manufacturer's website or the owners manual does not contain Peukert's exponent for the battery type!

However there is often a graph showing discharge time against discharge current. Or sometimes a table that shows available run times at different discharge currents. This information can be used to calculate Peukert's exponent for the battery type. All that is needed are two discharge times and discharge currents. These can either be lifted directly from the tables or from the graph.

Assume a battery data sheet shows a capacity of 100Ahrs at the 20 hour rate and 71.92Ahrs at the 4.794 rate. ie 100Ahrs @ 20hrs and 71.92Ahrs @ 4.794hrs. We have chosen these values as we worked them out before so it makes checking the derived formula (to follow) easier. These figures equate to a battery rated at 100Ahrs@20 hours with a Peukert's exponent of 1.3

In the first case (100Ahrs) we know that I₁=C₁/R₁ and therefore we know that C_p1=C₁(C₁/R₁)^n-1

Similarly for the second case (71.92Ahrs) we know that I₂=C₂/R₂ and therefore C_p2=C₂(C₂/R₂)^n-1

But we also know that the Peukert Capacity C_p1 and C_p2 must be equal because this never changes for any one particular battery.

Therefore we write:-

C_p1 = C_p2

And therefore C₁(C₁/R₁)^n-1 = C₂(C₂/R₂)^n-1

Thus we may also write:-

Log [C₁(C₁/R₁)^n-1] = Log [C₂(C₂/R₂)^n-1]

This can be simplified to:-

Log C₁+(n-1)Log(C₁/R₁) = Log C₂+(n-1)Log(C₂/R₂)

Rearranging we get

(n-1)Log (C₁/R₁) - (n-1)Log (C₂/R₂) = Log C₂-Log C₁

Therefore:-

(n-1) [Log (C₁/R₁) - Log (C₂/R₂)] = Log C₂-Log C₁

This simplifies to

(n-1) [Log (C₁R₂/C₂R₁)] = Log (C₂/C₁)

Therefore:-

(n – 1) = [Log (C₂/C₁)] / [Log (C₁R₂/C₂R₁)]

Therefore

n = 1 + [Log (C₂/C₁)] / [Log (C₁R₂/C₂R₁)]

Which simplifies again to

n = [Log (R₂/R₁)] / [Log (C₁/R₁) – Log (C₂/R₂)]

Therefore (working through our example battery):-

n = [Log (4.794/20)] / [Log (100/20) - Log (71.92/4.794)] = 1.3 which we know to be correct as that is what we started with.

So from this equation, given any two discharge rates and times or any two capacities at different hour ratings we can calculate Peukert's exponent, which then makes it easy to calculate any other discharge times and/or capacities.

Now you may be asking yourself why we went to all this trouble to develop a seemingly very complicated formula involving logarithms when an apparently much simpler one will do the job. The answer is simple.....

We kept getting asked why our apparently simple formula worked perfectly and agreed with Peukert calculators found on the internet that seemed much more complicated than ours. We just wanted to show that they are actually exactly the same formula rearranged. Please note that there are many Peukert calculators on the internet that are wrong. They do not work. They use the original Peukert's equation [T=C/Iⁿ] without any correction for the battery discharge rating.

And finally we have incorporated this formula into a spreadsheet that allows you to enter two discharge times, two capacities and it will calculate Peukert's exponent to save you the trouble.

Many thanks to Chris Wyles BSc(hons) MBA for the mathematical proofs, algebra and proof reading without whose help this article would have taken somewhat longer to complete (like about never).

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